Wikipedia:Reference desk/Archives/Mathematics/2015 November 15

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November 15

piecewise polynomial least squares

(I tried math.stackexchange and got no response for a week. Boo hoo.)

I have in mind a project involving a least-squares fit using piecewise polynomials; at a finite number of known arguments x_j, the k_jth derivative is discontinuous.

How many basis functions are needed? My guess is: xⁿ for 0≤n<min(k), and then, for each j,n such that k_j ≤ n ≤ the maximum degree, a pair of functions which are zero on one side and (x-x_j)ⁿ on the other. Is that right?

In general, I welcome any pointers that might reduce the number of wheels I'll reinvent. —Tamfang (talk) 07:00, 15 November 2015 (UTC)

Just trying to understand the problem here. You are trying to create a spline composed of multiple polynomial arcs, right ? The adjacent arc endpoints must have point continuity, of course, but how about tangent & curvature continuity, etc. ? Since you are using least squares method, I assume you don't need an exact fit. So, how many points would each arc run through ? (Just offhand, this method sounds like it would generate an extremely "lumpy" spline.) I assume you already know how the number of constraints relates to the degree of the polynomial ? StuRat (talk) 07:12, 15 November 2015 (UTC)

Sure, let's say I'm trying to create a spline composed of multiple polynomial arcs, and the degree of continuity is k_j-1. Maybe I like it runny lumpy; if it's lumpier than I like, I'll increase k_j. Rather than discrete points, my input is piecewise continuous, so the algo involves integrals rather than sums. Number of constraints, in the sense I think you mean, is not meaningful here. —Tamfang (talk) 08:51, 15 November 2015 (UTC)

If the function is on the domain $[x_{0},x_{m}]$, and the polynomials are of degree at most d, and for $1\leq j<m$ the derivatives at $x_{j}$ are expected to be continuous up to $k_{j}-1$ ($k_{j}$ constraints), then I'm pretty sure the number of degrees of freedom is $m(d+1)-\sum _{j=1}^{m-1}k_{j}$. -- Meni Rosenfeld (talk) 09:53, 15 November 2015 (UTC)

And I think the following basis functions will work (probably the same as what you wrote, but I think is clearer): Letting $k_{0}=0$, for each $0\leq j<m$ and $k_{j}\leq n\leq d$, the function which is 0 for $x\leq x_{j}$ and $(x-x_{j})^{n}$ for $x>x_{j}$. This also means their number can be rewritten as $\sum _{j=0}^{m-1}(d+1-k_{j})$. -- Meni Rosenfeld (talk) 10:06, 15 November 2015 (UTC)

Hm ... thanks, yes, I think that does work; the k₀=0 is a good gimmick (removing some special cases from the description). You've saved me some redundancy; I was thinking that for each discontinuity I'd need pairs of functions: zero on the left and (x-x_j)ⁿ on the right, (x-x_j)ⁿ on the left and zero on the right. —Tamfang (talk) 03:51, 24 November 2015 (UTC)

Meni, thanks again; see result at [1]. —Tamfang (talk) 22:51, 8 December 2015 (UTC)

This is above my head (I don't know why I even look at this notice board!) but does Savitzky–Golay filter help? Thincat (talk) 09:00, 20 November 2015 (UTC)

That's interesting, but no. —Tamfang (talk) 03:03, 22 November 2015 (UTC)